[M] Let \(H = {\bf{span}}\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\) and \({\rm B} = \left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}}} \right\}\). Show that \(x\) is in H, and find the \(\beta - \)coordinate vector of x, when

\({{\bf{v}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}{{\bf{11}}}\\{ - {\bf{5}}}\\{{\bf{10}}}\\{\bf{7}}\end{array}} \right]\), \({{\bf{v}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{{\bf{14}}}\\{ - {\bf{8}}}\\{{\bf{13}}}\\{{\bf{10}}}\end{array}} \right]\), \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{\bf{19}}}\\{ - {\bf{13}}}\\{{\bf{18}}}\\{{\bf{15}}}\end{array}} \right]\)

The matrix formed using thecolumn vectors is

\(A = \left[ {\begin{array}{*{20}{c}}{11}&{14}&{19}\\{ - 5}&{ - 8}&{ - 13}\\{10}&{13}&{18}\\7&{10}&{15}\end{array}} \right]\).

Consider matrix \(A = \left[ {\begin{array}{*{20}{c}}{11}&{14}&{19}\\{ - 5}&{ - 8}&{ - 13}\\{10}&{13}&{18}\\7&{10}&{15}\end{array}} \right]\).

Use the code in MATLAB to obtain therow-reducedechelon form as shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {{\rm{ }}\begin{array}{*{20}{c}}{11}&{14}&{19;\,\,\begin{array}{*{20}{c}}{ - 5}&{ - 8}&{ - 13;\,\,\begin{array}{*{20}{c}}{10}&{13}&{18;\;\begin{array}{*{20}{c}}7&{10}&{15}\end{array}}\end{array}}\end{array}}\end{array}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\(\left[ {\begin{array}{*{20}{c}}{11}&{14}&{19}\\{ - 5}&{ - 8}&{ - 13}\\{10}&{13}&{18}\\7&{10}&{15}\end{array}} \right] \sim \left[{\begin{array}{*{20}{c}}1&0&{ - \frac{5}{3}}\\0&1&{\frac{8}{3}}\\0&0&0\\0&0&0\end{array}} \right]\)

The equation \({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} = {\bf{x}}\) is consistent.The values of \({c_1}\) and \({c_2}\) from the echelon form are

\({c_1} = - \frac{5}{3}\)and \({c_2} = \frac{8}{3}\).

So, the \(\beta - \)coordinates of \({\bf{x}}\) are \(\left( { - \frac{5}{3},\frac{8}{3}} \right)\).